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12 July, 05:24

What is the molality of a solution prepared by dissolving 75.8 g of ethylene glycol, HOCH2CH2OH, in 1.55 L of water? Assume the density of water is 1.00 g/mL.

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  1. 12 July, 07:03
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    0.787 mol. Kg⁻¹ or 0.787 m

    Explanation:

    Data Given:

    Mass = m = 75.8 g

    Volume = V = 1.55 L = 1550 mL

    M. Mass of Ethylene Glycol = 62.07 g/mol

    Density of water = d = 1.0 g/mL

    To Find:

    Molality = ?

    Molality is a unit of concentration and is defined as the number of moles of solute per Kg of a solvent.

    Molality = Moles of Solute / Kg of Solvent - - - (1)

    First Calculate Moles of Solute as;

    Moles = Mass / M. Mass

    Moles = 75.8 g / 62.07 g/mol

    Moles = 1.22 moles

    Secondly calculate mass of water as,

    mass of water = density * volume

    mass of water = 1.0 g/mL * 1550 mL

    mass of water = 1550 g or 1.55 Kg

    Now, putting values of moles and mass of solvent in equation 1,

    Molality = 1.22 mol / 1.55 Kg

    Molality = 0.787 mol. Kg⁻¹ or 0.787 m
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